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in Lines and Angles by (10.9k points)
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Find X in the following figure

3 Answers

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by (10.9k points)
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Best answer

Using  sine law of properties of triangle we get for triangles BCD and ACD

\({sin(78+Y)\over sinX }={a\over b} \)  and \({sin78\over sin24}={a\over b}\)

It is obvious from the figure that  X+Y=72, So Y = 72-X

Hence \({sin(78+Y)\over sinX}= {sin78\over sin24}\)

=>\({sin(150-X)\over sinX}={sin(90-12)\over sin24}\)

=>\({sin150cosX-cos150sinX\over sinx}={cos12\over 2sin12cos12}\)

=>\({sin30cosX\over sinX}+cos30^o={1\over 2sin12}\)

=>\({cotX \over 2}+{\sqrt3\over 2}={cosec 12\over 2}\)

=> \(X=arctan{1\over cosec12-\sqrt3}=18^o\)

calculator used

Please see the link below in support of this solution

https://www.sarthaks.com/3174373/prove-the-identity?show=3179819#a3179819

by (41.0k points)
+1
your solution is correct and I learnt a new and good method of solution of such type of question. Thanks.
by (10.9k points)
Thank you
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by (42.1k points)

∠1 + 24° + 48° = 180°

⇒ ∠1 = 108°

Also ∠1 + ∠2 = 180°

⇒ ∠2 = 180° - 108° = 72°

∵ External angle is equal to its opposite interior angle.

∴ x + y = ∠2

⇒ x + y = 72°

We only can get a relation with the help of given information not the exactly value of x.

If there is options given or other information given then may get an answer but recently I am unable to get it.

by (10.9k points)
edited by
it is obvious from figure that x+y=72deg.
The another information given here is equality of three sides ,which is not taken into account..
0 votes
by (41.0k points)

No, we need some other information like if this quadrilateral is part of any circle or some other way in which we can differentiate x and y But if we use only these three equal sides then we only get x + y = 72°(in all cases)

In isosceles triangle ABC,

∠B = ∠C = \(\frac{180^{\circ}-24^{\circ}}2\) = \(\frac{156^{\circ}}2=78^{\circ}\)

∴ ∠5 = 78° - 48° = 30°
∠6 = ∠C = 78°

In Δ BCD, ∠5 + ∠6 + y + x = 180°

⇒ x + y = 180° - 30° - 78° = 72°

In isosceles triangle ABD

∠A + ∠4 = ∠3 = \(\frac{180^{\circ}-48^{\circ}}2=\frac{132^{\circ}}2=60^{\circ}\)

∴ ∠3 = 66°

∠1 = 66° - ∠A = 66° - 24° = 42°

with its help we also can get (in Δ ACD)

x + y = 72°

But not a separate value of x & y.

  ∵ x + y = 72° and x < y (∵ CE < DE)

∴ x < 36° 

(If options are given then can apply it otherwise did not get the logic which solve it )

by (10.9k points)
I get an answer which is given here , please check

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